Sketch the curve y=4-(x+3)^2, showing the points where the curve crosses the x-axis and any minimum or maximum points.

This equation rearranges to give -y=(x+3)^2-4, which is very similar to our curve y=(x+3)^2-4 from before. In fact, replacing y with -y in an equation is equivalent to reflecting the curve through the x-axis. We then take the points (-5,0), (-1,0) and (-3,-4) from before and replace y with -y, giving (-5,0), (-1,0) and (-3,4). We have found where the new curve crosses the x-axis and its minimum/maximum. The graph is an inverted u-shape since we have a -x^2 in the equation so (-3,4) is a maximum point.

JI

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